$A=(a_{ij})$ is an $n\times n$ matrix and $\det(A)=d\neq 0$. Removing the last line and the last column of $\text{adj}(A),$ we get the $(n-1) \times (n-1)$ matrix $B$.
Find $\det(B)$.
Thanks.
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1$\begingroup$ Looks like the answer can be obtained by a theorem of Jacobi which can be found on page 98 here: maths.ed.ac.uk/~aar/papers/aitken.pdf $\endgroup$– CasteelsOct 29, 2013 at 11:22
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1$\begingroup$ If $n=2$, doesn't $\det(B)$ just turn out to be the lower right entry of $A$? How can you know what that is, if all you know is $\det A$? $\endgroup$– Gerry MyersonOct 29, 2013 at 11:32
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$\begingroup$ I was thinking the same thing. It's not clear whether or not $A$ is given. $\endgroup$– CasteelsOct 29, 2013 at 11:35
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$\begingroup$ Observing Casteels's comment, it seems to be we'd get that $$\det B= A_{nn}\cdot\left(\det A\right)^{n-1}$$ $\endgroup$– DonAntonioOct 29, 2013 at 12:23
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$\begingroup$ Hi,@DonAntonio,this seems not to work for $A=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \\ \end{array} \right)$ $\endgroup$– Yuxuan DongOct 29, 2013 at 12:36
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2 Answers
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$\det(B)$ is just the right-bottom entry of $adj(adj(A))$. And $adj(adj(A))=d^{n-2}A$.So $\det(B)=d^{n-2}A_{n,n}$.
I learned it from somebody just now.
And in another hand,we can use Jacobi's theorem.(see comments)
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Here is the algebraic derivation. First write $$\bf{A} = \left(\matrix{\bf{M} & \bf{b} \\ \bf{c}^\top & k}\right) \,\,\,\text{and}\,\,\,\, \bf{A}^{-1}=\pmatrix{\bf{E} & \bf{f} \\ \bf{g}^\top & h}$$ and define determinant $\Delta = \det\bf{A}$. The dimension of $\bf{A}$ is $n$. The question then is the determinant of the matrix $\left(\Delta\bf{I}\cdot\bf{E}\right)$ which has dimension $n-1$. Let $k$ be non-zero (otherwise the result is trivial; multiplying the terms of $A^{-1}A$ shows $E$ to be singular).
Rewriting the determinant: $$\Delta = \det\left[\overbrace{\pmatrix{\bf{I} & -\mathbf{b}\frac{1}{k} \\ \mathbf{0}^\top & 1\\}}^{\det = 1}\pmatrix{\bf{M} & \bf{b} \\ \bf{c}^\top & k}\right]$$ or $$\Delta = \det\left(\mathbf{M} - \mathbf{b}\frac{1}{k}\mathbf{c}^\top\right)\cdot k$$ Since the $\bf{b}$ column was zeroed out, the determinant is the multiplication of the blocks' determinants. Also the unimodular row operation matrix does not change the determinant (standard algebra).
Now with that form of the determinant handy, we focus on the $\bf{E}$ matrix. Using a previously derived formula (also worked out algebraically) we write: $$\mathbf{E}^{-1} = \mathbf{M} - \mathbf{b}\frac{1}{k}\mathbf{c}^\top$$
Hmm, that looks familiar, it is the matrix that determined $\Delta$. So now we have
$$\Delta = \det\left(\mathbf{E}^{-1}\right)\cdot k$$ or $$\frac{1}{\det\left(\mathbf{E}^{-1}\right)}= \det(\mathbf{E}) = \frac{k}{\Delta}$$
And we finally arrive at the formula of interest (so long as we remember that the dimension of $\mathbf{E}$ is $n-1$)
$$\det\left(\Delta\bf{I}\cdot\bf{E}\right) = \Delta^{n-1}\frac{k}{\Delta} = \Delta^{n-2}k$$
